Problem 47 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\pi / 8} \cos 2 x d x$$

Step-by-Step Solution

Answer

Answer: The value of the integral is $$\frac{\sqrt{2}}{4}$$.

Step 1: Find the antiderivative of the function

First, let's find the antiderivative of the given function, which is $$\cos 2x$$. To do this, we will apply integration by substitution. Let $$u = 2x$$. Then, $$d u = 2dx$$. Hence, the given integral can be rewritten as: $$\int \cos 2x dx = \frac{1}{2} \int \cos u du$$ Now, we know that the antiderivative of the $$\cos u$$ function is $$\sin u$$. Therefore, $$\frac{1}{2}\int \cos u du = \frac{1}{2}\sin u + C$$ Now, substitute the original value of u, which is $$2x$$: $$\frac{1}{2}\sin u + C = \frac{1}{2}\sin {(2x)} + C$$ So, the antiderivative of the given function is $$\frac{1}{2}\sin {(2x)} + C$$.

Step 2: Apply the Fundamental Theorem of Calculus

Now, we will apply the Fundamental Theorem of Calculus to evaluate the given integral. Since we have found the antiderivative of the function, we can use the theorem to find the integral: $$\int_{0}^{\pi/8} \cos 2x dx = \left[\frac{1}{2}\sin{(2x)}\right]_{0}^{\pi/8}$$ Evaluate the expression at the upper limit and at the lower limit: $$\left[\frac{1}{2}\sin{(2x)}\right]_{0}^{\pi/8} = \frac{1}{2}\sin{\left(2 \cdot \frac{\pi}{8}\right)} - \frac{1}{2}\sin{(2 \cdot 0)}$$ After simplification, we get: $$\frac{1}{2}\sin{\left(\frac{\pi}{4}\right)} - \frac{1}{2}\sin{0} = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - 0$$ Finally, after simplification, we get: $$\frac{\sqrt{2}}{4}$$ Thus, the value of the integral $$\int_{0}^{\pi/8} \cos 2x dx$$ is $$\frac{\sqrt{2}}{4}$$

Key Concepts

Integration by Substitution

Integration by substitution is a crucial technique used to simplify solving integrals. The process involves changing the variable of integration to make the integral easier to evaluate. In the given problem, we started by identifying that the function inside the integral was \[ \cos{2x} \]. To make it more manageable, we set \( u = 2x \), where \( du = 2 \, dx \). This transformation allows the integral to be expressed in terms of \( u \) instead of \( x \).
This lets us rewrite the integral as \( \int \cos{2x} \, dx = \frac{1}{2} \int \cos{u} \, du \). This substitution simplifies the integral to a basic form involving the cosine function. Therefore, through substitution, one takes a seemingly complex integral and transforms it into one that is straightforward to solve. This method is universally applicable and is pivotal in calculus for solving more complex integrals.

Definite Integrals

Definite integrals help us to calculate the accumulated value of a function over a specific interval, which in our example is from \( 0 \) to \( \frac{\pi}{8} \). When we calculate a definite integral, our goal is to determine the net area between the function and the x-axis over this interval.
By using the antiderivative obtained through our previous substitution, \( \frac{1}{2}\sin{(2x)} + C \), we apply the Fundamental Theorem of Calculus. This theorem allows us to utilize the antiderivative to find the exact value of the integral over the interval:

• Evaluate the antiderivative at the upper limit: \( \frac{1}{2}\sin{\left( \frac{\pi}{4} \right)} \)
• Evaluate at the lower limit: \( \frac{1}{2}\sin{0} \)
• Subtract the lower limit evaluation from the upper limit evaluation

The result \( \frac{\sqrt{2}}{4} \) gives the area under the curve between \( 0 \) and \( \frac{\pi}{8} \). Thus, definite integrals allow us to perform precise calculations of areas and accumulated values over specific ranges.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. In simpler terms, it is the reverse process of differentiation. Finding an antiderivative is crucial since it helps in evaluating definite integrals.
For the function \( \cos{2x} \), we determined the antiderivative through integration by substitution. Converting the variable allowed us to integrate \( \cos{u} \), resulting in \( \sin{u} \). Rewriting this in terms of \( x \) gave us \( \frac{1}{2}\sin{(2x)} + C \).

• The constant \( C \) is added because the derivative of a constant is zero, meaning any constant could have been lost during differentiation.

The antiderivative is essential in evaluating definite integrals because it provides a specific expression that, when evaluated at the bounds, gives the result of the integral. Thus, understanding antiderivatives is foundational for calculus, making the solving of integrals possible through reverse differentiation.